A point object 'O' is kept in a medium of refractive index $n_1$ in front of a convex spherical surface of radius of curvature R which separates the second medium of refractive index $n_2$ from the first one, as shown in the figure.
- Draw the ray diagram showing the image formation and deduce the relationship between the object distance and the image distance in terms of $n_1, n_2$ and R.
- When the image formed above acts as a virtual object for concave spherical, surface separating the medium $n_2$ from $n_1 (n_2> n_1)$, draw this ray diagram and write the similar (similar to (a) relation. Hence obtain the expression for the lens maker's formula.
CBSE DELHI - SET 3 2015
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For small angles
$\angle\text{N0M}\cong\tan\angle \text{N0M} = \frac{\text{MN}}{\text{OM}}$
$\angle\text{NCM}\cong\tan\angle \text{NCM} = \frac{\text{MN}}{\text{MC}}$
$\angle\text{NIM}\cong\tan\angle \text{NIM} = \frac{\text{MN}}{\text{MI}}$
In $\Delta\text{NOC} ,$ $\angle \text{i} = \angle\text{i}\angle\text{NOM} + \angle\text{NCM}$
$\therefore\angle\text{i} = \frac{\text{MN}}{\text{OM}} + \frac{\text{MN}}{\text{MC}}$
$\angle\text{r} = \angle \text{NCM} -\angle \text{NIM}$
$=\frac{\text{MN}}{\text{MC}} -\frac{\text{MN}}{\text{MI}}$
Using Snell's Law
$n_1 \sin\ i. = n_2 \sin\ r$
For small angles $n_1\ i = n_2\ r$
Substituting for i and r, we get
$\frac{\text{n}_{1}}{\text{OM}} +\frac{\text{n}_{2}}{\text{MI}} = \frac{\text{n}_{2} -\text{n}_{1}}{\text{MC}}$
Here, $\text{OM} =\text{-u}, \text{MI} = + \text{v},\text{MC} = + \text{R}$
Substituting these, we get
$\Rightarrow\frac{\text{n}_{2}}{\text{v}} - \frac{\text{n}_{1}}{\text{u}} = \frac{\text{n}_{2} -\text{n}_{1}}{\text{R}}$
(Alternatively accept this Ray diagram)
Similarly relation for the surface ADC.
$\frac{-\text{n}}{\text{DI}{1}} +\frac{\text{n}_{1}}{\text{DI}} = \frac{\text{n}_{2} - \text{n}_{1}}{\text{DC}_{2}}$ ................................(i)
Refraction at the first surface ABC of the lens.
$\frac{\text{n}_{1}}{\text{OB}} +\frac{\text{n}_{2}}{\text{BI}_{1}} = \frac{\text{n}_{2} -\text{n}_{2}}{\text{BC}_{1}}$..............................(ii)
Adding (i)and (ii), and taking $BI_1\cong_DI_1, $ we get
$\frac{\text{n}_{1}}{\text{OB}} +\frac{\text{n}_{1}}{\text{DI}} = (\text{n}_{2} -\text{n}_{1})\big(\frac{1}{\text{BC}_{1}} + \frac{1}{\text{DC}_{2}}\big)$
Here, $OB = – u$
$DI = + v$
$BC_1 = + R_1$
$DC_2 = – R_2$
$\Rightarrow = \frac{\text{n}_{1}}{\text{-u}} + \frac{\text{n}_{1}}{\text{v}} = (\text{n}_{2} -\text{n}_{1})\bigg(\frac{1}{\text{R}_{1}} +\frac{1}{\text{R}_{2}}\bigg)$
$\Rightarrow = \text{n}_{1}\bigg(\frac{1}{\text{v}} +\frac{1}{u}\bigg) (\text{n}_{2} -\text{n}_{1})\bigg(\frac{1}{\text{R}_{1}} +\frac{1}{\text{R}_{2}}\bigg)$
$\Rightarrow\frac{1}{\text{f}} =\bigg(\frac{\text{n}_{2}}{\text{n}_{1}} -1 \bigg)\bigg(\frac{1}{\text{R}_{1}} -\frac{1}{\text{R}_{2}}\bigg)$
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